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Topic: NBA teams (Read 1861 times) |
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Altamira_64
Junior Member
Posts: 116
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We have 20 NBA players and want to create 4 teams of 5 players each. How many different ways are there?
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Altamira_64
Junior Member
Posts: 116
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Re: NBA teams
« Reply #1 on: Nov 8th, 2012, 9:14am » |
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I thought it was 20!/15! but apparently it isn't...
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Grimbal
wu::riddles Moderator Uberpuzzler
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Re: NBA teams
« Reply #2 on: Nov 8th, 2012, 9:31am » |
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Maybe 20!/(5!^4)/4! = 488864376
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rmsgrey
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Re: NBA teams
« Reply #3 on: Nov 9th, 2012, 4:25am » |
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on Nov 8th, 2012, 9:31am, Grimbal wrote:Maybe 20!/(5!^4)/4! = 488864376 |
| I agree with the formula (I haven't checked the arithmetic) Take your 20 players and line them up (20! possibilities) Take the first five for team A, the next five for team B, 11-15 for team C and the remaining five are team D. There are 5! ways of ordering the players on each time, so 5!4 ways of getting the same four teams assigned in the same order, but if you swap the labels of the four teams around, you end up with the same actual teams, just in a different order, and there are 4! ways of ordering the four teams. So there are 5!4*4! ways of getting each (unordered) set of teams out of the original 20! orders of the players. If it makes a difference which team is which, then it's 20!/(5!4) possible teams; if the teams are equivalent apart from their players, then it's Grimbal's figure.
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pex
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Re: NBA teams
« Reply #4 on: Nov 9th, 2012, 7:11am » |
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Yet another way to get the same result: - pick five people for Team A (20!/(5!15!) possibilities) - from the remaining fifteen, pick five for Team B (15!/(5!10!)) - again for Team C (10!/(5!5!)) - Team D is fixed now, because only five people are left. Overall, 20!/(5!15!) * 15!/(5!10!) * 10!/(5!5!) = 20!/(5!4) possibilities if it matters which team has which name. If not, divide by another factor of 4!.
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